Mixing
Sum of geometric series where $x=sin^{-1}{frac{7}{8}}$
$begingroup$
This problem is from a local contest.
The series $sin{x}$, $sin{2x}$, $4sin{x}-4sin^3{x}$, $...$ is a geometric series if $x=sin^{-1}{frac{7}{8}}$. Compute the sum of the geometric series.
I did not compute the solution in time but this was my reasoning.
For the first term, $sin{sin^{-1}{frac{7}{8}}}=frac{7}{8}$
I was not able to find $sin{2x}$. I did not attempt the double angle identity since that would have introduced cosine.
For the third term, I found $4sin{sin^{-1}{frac{7}{8}}}=frac{28}{8}$ and $4sin^3{sin^{-1}{frac{7}{8}}}=4(frac{7}{8})^3=frac{1372}{512}$. The third term is the difference of $frac{28}{8}$ and $frac{1372}{512}$.
My question is, how can I compute the second term, $sin{2x}$?
trigonometry contest-math
asked Jan 17 at 21:15
GnumbertesterGnumbertester
623113
$endgroup$
add a comment |
$begingroup$
This problem is from a local contest.
The series $sin{x}$, $sin{2x}$, $4sin{x}-4sin^3{x}$, $...$ is a geometric series if $x=sin^{-1}{frac{7}{8}}$. Compute the sum of the geometric series.
I did not compute the solution in time but this was my reasoning.
For the first term, $sin{sin^{-1}{frac{7}{8}}}=frac{7}{8}$
I was not able to find $sin{2x}$. I did not attempt the double angle identity since that would have introduced cosine.
For the third term, I found $4sin{sin^{-1}{frac{7}{8}}}=frac{28}{8}$ and $4sin^3{sin^{-1}{frac{7}{8}}}=4(frac{7}{8})^3=frac{1372}{512}$. The third term is the difference of $frac{28}{8}$ and $frac{1372}{512}$.
My question is, how can I compute the second term, $sin{2x}$?
trigonometry contest-math
asked Jan 17 at 21:15
GnumbertesterGnumbertester
623113
$endgroup$
1
$begingroup$
The first term is $sin(x)$ and subsequent terms multiply by $2cos(x)$. Can you finish now?
$endgroup$
– Mark Viola
Jan 17 at 21:25
1
$begingroup$
Use $sin2x=2sin xcos x=2sin xsqrt{1-sin^2x}$.
$endgroup$
– Yves Daoust
Jan 17 at 21:46
add a comment |
$begingroup$
This problem is from a local contest.
The series $sin{x}$, $sin{2x}$, $4sin{x}-4sin^3{x}$, $...$ is a geometric series if $x=sin^{-1}{frac{7}{8}}$. Compute the sum of the geometric series.
I did not compute the solution in time but this was my reasoning.
For the first term, $sin{sin^{-1}{frac{7}{8}}}=frac{7}{8}$
I was not able to find $sin{2x}$. I did not attempt the double angle identity since that would have introduced cosine.
For the third term, I found $4sin{sin^{-1}{frac{7}{8}}}=frac{28}{8}$ and $4sin^3{sin^{-1}{frac{7}{8}}}=4(frac{7}{8})^3=frac{1372}{512}$. The third term is the difference of $frac{28}{8}$ and $frac{1372}{512}$.
My question is, how can I compute the second term, $sin{2x}$?
trigonometry contest-math
asked Jan 17 at 21:15
GnumbertesterGnumbertester
623113
$endgroup$
This problem is from a local contest.
The series $sin{x}$, $sin{2x}$, $4sin{x}-4sin^3{x}$, $...$ is a geometric series if $x=sin^{-1}{frac{7}{8}}$. Compute the sum of the geometric series.
I did not compute the solution in time but this was my reasoning.
For the first term, $sin{sin^{-1}{frac{7}{8}}}=frac{7}{8}$
I was not able to find $sin{2x}$. I did not attempt the double angle identity since that would have introduced cosine.
For the third term, I found $4sin{sin^{-1}{frac{7}{8}}}=frac{28}{8}$ and $4sin^3{sin^{-1}{frac{7}{8}}}=4(frac{7}{8})^3=frac{1372}{512}$. The third term is the difference of $frac{28}{8}$ and $frac{1372}{512}$.
My question is, how can I compute the second term, $sin{2x}$?
trigonometry contest-math
trigonometry contest-math
asked Jan 17 at 21:15
GnumbertesterGnumbertester
623113
asked Jan 17 at 21:15
GnumbertesterGnumbertester
623113
asked Jan 17 at 21:15
GnumbertesterGnumbertester
623113
asked Jan 17 at 21:15
GnumbertesterGnumbertester
623113
asked Jan 17 at 21:15
GnumbertesterGnumbertester
623113
623113
1
$begingroup$
The first term is $sin(x)$ and subsequent terms multiply by $2cos(x)$. Can you finish now?
$endgroup$
– Mark Viola
Jan 17 at 21:25
1
$begingroup$
Use $sin2x=2sin xcos x=2sin xsqrt{1-sin^2x}$.
$endgroup$
– Yves Daoust
Jan 17 at 21:46
add a comment |
1
$begingroup$
The first term is $sin(x)$ and subsequent terms multiply by $2cos(x)$. Can you finish now?
$endgroup$
– Mark Viola
Jan 17 at 21:25
1
$begingroup$
Use $sin2x=2sin xcos x=2sin xsqrt{1-sin^2x}$.
$endgroup$
– Yves Daoust
Jan 17 at 21:46
1
1
$begingroup$
The first term is $sin(x)$ and subsequent terms multiply by $2cos(x)$. Can you finish now?
$endgroup$
– Mark Viola
Jan 17 at 21:25
$begingroup$
The first term is $sin(x)$ and subsequent terms multiply by $2cos(x)$. Can you finish now?
$endgroup$
– Mark Viola
Jan 17 at 21:25
1
1
$begingroup$
Use $sin2x=2sin xcos x=2sin xsqrt{1-sin^2x}$.
$endgroup$
– Yves Daoust
Jan 17 at 21:46
$begingroup$
Use $sin2x=2sin xcos x=2sin xsqrt{1-sin^2x}$.
$endgroup$
– Yves Daoust
Jan 17 at 21:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A sequence $(a_n)$ is a geometric series if $a_{n+1}/a_n=c$, for every $n$. Thus the first two terms suffice to determine it:
$$
a_0=sin x,qquad a_1=sin2x=2sin xcos x
$$
Then
$$
c=frac{a_1}{a_0}=2cos x
$$
and indeed
$$
a_2=4sin x-4sin^3x=4sin xcos^2x=a_1cdot 2cos x
$$
Thus you have
$$
a_n=2^ncos^nxsin x
$$
and the sum of a geometric series converges if and only if $|c|<1$.
In your case
$$
c=2cos x=2sqrt{1-frac{49}{64}}=frac{sqrt{15}}{4}<1
$$
answered Jan 17 at 21:41
egregegreg
183k1486204
$endgroup$
add a comment |
$begingroup$
The series seems to be
$$S=sin(x)sum_{n=0}^{infty}(2cos(x))^n$$
Since $x=arcsin(7/8)$, $2cos(x)=sqrt{15}/4<1$. So the series converges and
$$S=(7/8)frac1{1-sqrt{15}/4}$$
answered Jan 17 at 21:29
Mark ViolaMark Viola
132k1277174
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A sequence $(a_n)$ is a geometric series if $a_{n+1}/a_n=c$, for every $n$. Thus the first two terms suffice to determine it:
$$
a_0=sin x,qquad a_1=sin2x=2sin xcos x
$$
Then
$$
c=frac{a_1}{a_0}=2cos x
$$
and indeed
$$
a_2=4sin x-4sin^3x=4sin xcos^2x=a_1cdot 2cos x
$$
Thus you have
$$
a_n=2^ncos^nxsin x
$$
and the sum of a geometric series converges if and only if $|c|<1$.
In your case
$$
c=2cos x=2sqrt{1-frac{49}{64}}=frac{sqrt{15}}{4}<1
$$
answered Jan 17 at 21:41
egregegreg
183k1486204
$endgroup$
add a comment |
$begingroup$
A sequence $(a_n)$ is a geometric series if $a_{n+1}/a_n=c$, for every $n$. Thus the first two terms suffice to determine it:
$$
a_0=sin x,qquad a_1=sin2x=2sin xcos x
$$
Then
$$
c=frac{a_1}{a_0}=2cos x
$$
and indeed
$$
a_2=4sin x-4sin^3x=4sin xcos^2x=a_1cdot 2cos x
$$
Thus you have
$$
a_n=2^ncos^nxsin x
$$
and the sum of a geometric series converges if and only if $|c|<1$.
In your case
$$
c=2cos x=2sqrt{1-frac{49}{64}}=frac{sqrt{15}}{4}<1
$$
answered Jan 17 at 21:41
egregegreg
183k1486204
$endgroup$
add a comment |
$begingroup$
A sequence $(a_n)$ is a geometric series if $a_{n+1}/a_n=c$, for every $n$. Thus the first two terms suffice to determine it:
$$
a_0=sin x,qquad a_1=sin2x=2sin xcos x
$$
Then
$$
c=frac{a_1}{a_0}=2cos x
$$
and indeed
$$
a_2=4sin x-4sin^3x=4sin xcos^2x=a_1cdot 2cos x
$$
Thus you have
$$
a_n=2^ncos^nxsin x
$$
and the sum of a geometric series converges if and only if $|c|<1$.
In your case
$$
c=2cos x=2sqrt{1-frac{49}{64}}=frac{sqrt{15}}{4}<1
$$
answered Jan 17 at 21:41
egregegreg
183k1486204
$endgroup$
A sequence $(a_n)$ is a geometric series if $a_{n+1}/a_n=c$, for every $n$. Thus the first two terms suffice to determine it:
$$
a_0=sin x,qquad a_1=sin2x=2sin xcos x
$$
Then
$$
c=frac{a_1}{a_0}=2cos x
$$
and indeed
$$
a_2=4sin x-4sin^3x=4sin xcos^2x=a_1cdot 2cos x
$$
Thus you have
$$
a_n=2^ncos^nxsin x
$$
and the sum of a geometric series converges if and only if $|c|<1$.
In your case
$$
c=2cos x=2sqrt{1-frac{49}{64}}=frac{sqrt{15}}{4}<1
$$
answered Jan 17 at 21:41
egregegreg
183k1486204
answered Jan 17 at 21:41
egregegreg
183k1486204
answered Jan 17 at 21:41
egregegreg
183k1486204
answered Jan 17 at 21:41
egregegreg
183k1486204
183k1486204
add a comment |
add a comment |
$begingroup$
The series seems to be
$$S=sin(x)sum_{n=0}^{infty}(2cos(x))^n$$
Since $x=arcsin(7/8)$, $2cos(x)=sqrt{15}/4<1$. So the series converges and
$$S=(7/8)frac1{1-sqrt{15}/4}$$
answered Jan 17 at 21:29
Mark ViolaMark Viola
132k1277174
$endgroup$
add a comment |
$begingroup$
The series seems to be
$$S=sin(x)sum_{n=0}^{infty}(2cos(x))^n$$
Since $x=arcsin(7/8)$, $2cos(x)=sqrt{15}/4<1$. So the series converges and
$$S=(7/8)frac1{1-sqrt{15}/4}$$
answered Jan 17 at 21:29
Mark ViolaMark Viola
132k1277174
$endgroup$
add a comment |
$begingroup$
The series seems to be
$$S=sin(x)sum_{n=0}^{infty}(2cos(x))^n$$
Since $x=arcsin(7/8)$, $2cos(x)=sqrt{15}/4<1$. So the series converges and
$$S=(7/8)frac1{1-sqrt{15}/4}$$
answered Jan 17 at 21:29
Mark ViolaMark Viola
132k1277174
$endgroup$
The series seems to be
$$S=sin(x)sum_{n=0}^{infty}(2cos(x))^n$$
Since $x=arcsin(7/8)$, $2cos(x)=sqrt{15}/4<1$. So the series converges and
$$S=(7/8)frac1{1-sqrt{15}/4}$$
answered Jan 17 at 21:29
Mark ViolaMark Viola
132k1277174
answered Jan 17 at 21:29
Mark ViolaMark Viola
132k1277174
answered Jan 17 at 21:29
Mark ViolaMark Viola
132k1277174
answered Jan 17 at 21:29
Mark ViolaMark Viola
132k1277174
132k1277174
add a comment |
add a comment |
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1
$begingroup$
The first term is $sin(x)$ and subsequent terms multiply by $2cos(x)$. Can you finish now?
$endgroup$
– Mark Viola
Jan 17 at 21:25
1
$begingroup$
Use $sin2x=2sin xcos x=2sin xsqrt{1-sin^2x}$.
$endgroup$
– Yves Daoust
Jan 17 at 21:46